gor@cs.strath.ac.uk (Gordon Russell) (02/11/91)
Hi out there,
I am hoping that someone on the network can answer a compiler implementation
problem for me. The problem stems from my PhD research into compiler
techniques. If we consider the following program extract...
variable1 : integer;
procedure MAIN is
variable1 : integer;
result : integer;
----------------------------------------------------
procedure GET_RESULT(variable2: in out integer) is
begin
variable1=2;
variable2=variable1 + variable2;
end GET_RESULT;
----------------------------------------------------
begin
variable1=10;
get_result(variable1);
end MAIN;
My question is.......what is variable1 equal to at the end of MAIN?
There appears to be a number of options....either
(1) 4
(2) 12
(3) Something wierd
(4) Compiler dependent.
Evidentally, it is reliant on whether GET_RESULT operates on variable2
directly or indirectly. I am hoping that the results are compiler
dependent. Does anyone have an ADA compiler who is willing to test this?
I am especially interested to hear from official sources (if they are
reading this), since I do not want to break any validation suite
program.
Reply either to this newsgroup or by email. I will post a concensus if
comments are mailed directly to me. Please no flames if this program is
not syntatically correct, as it is the mechanism which I am inquiring
after. And yes, I do think that this is poor programming practice,
but when has that stopped anyone!
Thanx in advance.....Gordon Russell
gor@cs.strath.ac.ukwayne@inmet.inmet.com (02/13/91)
The compiler actions are well defined. According to the LRM (section 6.2:6) for a scalar parameter (integers, etc) of mode IN OUT the value of the formal parameter, variable2, is a copy of the actual parameter, variable1 and it in this case it should have the value of 10. At the end of GET_RESULT variable1 will be updated from variable2. Mode IN OUT is not "call by reference" for scalar types. 1 procedure MAIN is 2 variable1 : integer; 3 result : integer; 4 ---------------------------------------------------- 5 procedure GET_RESULT(variable2: in out integer) is 6 begin 7 variable1 := 2; 8 variable2 := variable1 + variable2; 9 end GET_RESULT; 10 ---------------------------------------------------- 11 begin 12 variable1 := 10; 13 get_result(variable1); 14 end MAIN; So at line 6 both variable1 and variable2 have the value of 10. After line 7 is executed, variable1 has the value of 2, and variable2 still has the value of 10. The answer should therefore be (2) 12 Wayne Wylupski Intermetrics, Inc.